Question: Solve for $x$ and $y$ using substitution. ${6x+5y = 3}$ ${x = -4y+10}$
Since $x$ has already been solved for, substitute $-4y+10$ for $x$ in the first equation. ${6}{(-4y+10)}{+ 5y = 3}$ Simplify and solve for $y$ $-24y+60 + 5y = 3$ $-19y+60 = 3$ $-19y+60{-60} = 3{-60}$ $-19y = -57$ $\dfrac{-19y}{{-19}} = \dfrac{-57}{{-19}}$ ${y = 3}$ Now that you know ${y = 3}$ , plug it back into $\thinspace {x = -4y+10}\thinspace$ to find $x$ ${x = -4}{(3)}{ + 10}$ $x = -12 + 10$ ${x = -2}$ You can also plug ${y = 3}$ into $\thinspace {6x+5y = 3}\thinspace$ and get the same answer for $x$ : ${6x + 5}{(3)}{= 3}$ ${x = -2}$